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The Human Eye and Colourful World
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28/08/2021 6:55 pm
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The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).
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28/08/2021 6:59 pm
The person needs a convex lens to rectify this defect.
Calculation of power of the lens:
The hypermetropic eye can see the nearby object kept at 25 cm clearly if the image is formed at its own near point i.2. 50 cm.
So, object distance, u = -25 cm
Image distance, v = -50 cm
\(\frac{1}{v}\) - \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{-50}\) - \(\frac{1}{ -25}\) = \(\frac{1}{f}\)
f = 50 cm
P = \(\frac{100}{f}\) = \(\frac{100}{50}\)
= 2D